Question

Find a set of vectors {u⃗ ,v⃗ }{u→,v→} in r4r4 that spans the solution set of the equations {w−x+2y+4z5w+2x−y+3z==0,0. {w−x+2y+4z=0,5w+2x−y+3z=0.

Answer 1

Given

$w-x+2y+4z=0 \\ 5w+2x-y+3z=0$

We can rewrite it in matrix form as:

$\left[\begin{array}{cccc}1&-1&2&4\\5&2&-1&3\\0&0&0&0\\0&0&0&0\end{array}\right] \left[\begin{array}{c}w\\x\\y\\z\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right] \\ \\ \left[\begin{array}{cccc}1&-1&2&4\\5&2&-1&3\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ -5R_1+R_2\rightarrow R_2$
$\left[\begin{array}{cccc}1&-1&2&4\\0&7&-11&-17\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ \frac{1}{7} R_2\rightarrow R_2 \\ \\ \left[\begin{array}{cccc}1&-1&2&4\\0&1&-\frac{11}{7}&-\frac{17}{7}\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ R_1+R_2\rightarrow R_1$
$\\ \\ \left[\begin{array}{cccc}1&0&\frac{3}{7}&\frac{11}{7}\\0&1&-\frac{11}{7}&-\frac{17}{7}\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right] \\ \\ \Rightarrow w= -\frac{3}{7} y- \frac{11}{7} z \\ x=\frac{11}{7} y+ \frac{17}{7} z \\ y=free \\ z=free \\ \\ =y\left\ \textless \ -\frac{3}{7},\frac{11}{7}1,0\right\ \textgreater \ +z\left\ \textless \ - \frac{11}{7},\frac{17}{7},0,1\right\ \textgreater$

Thus, the solution set is a span of $\{\left\ \textless \ -\frac{3}{7},\frac{11}{7}1,0\right\ \textgreater \ ,\left\ \textless \ - \frac{11}{7},\frac{17}{7},0,1\right\ \textgreater \ \}$